Yesterday, someone asked whether an $n\times n$ row-stochastic matrix $A$ that satisfies $a_{ii}>\max(a_{ij},a_{ji})$ for all $i\ne j$ is necessarily nonsingular. The answer is clearly positive when $n=1,2$. I have shown that the answer is negative for every $n\ge4$ and the counterexample $A$ can be chosen to be symmetric (hence doubly stochastic) and entrywise positive.
This leaves only the case $n=3$ open. I have a feeling that $A$ must be nonsingular but I cannot prove it. Any idea?
Here is one of my discoveries so far. Suppose $Av=0$ has a nontrivial solution $v=(x,y,z)^T$. By permuting the rows and columns of $A$ and by scaling $v$ (by a possibly negative factor) if necessary, we may assume that either $x=1\ge y\ge z>-1$ or $x=1\ge y\ge0>z=-1$. Note that none of $y$ and $z$ can be zero, otherwise some column of $A$ is a scalar multiple of another, but this is a contradiction to the assumption that $a_{ii}>\max(a_{ij},a_{ji})$. So, either $y>0>z$ or $y,z<0$. The former case is impossible, because it implies that $$ a_{11}<a_{11}+a_{12}y=a_{13}|z|\le a_{13}. $$ Hence we must have $y,z<0$. In turn, we obtain $x=1>0>y\ge z>-1$, but I cannot proceed any further.
Lemma: If $x, y, a_1, a_2, b_1, b_2$ are non-negative terms such that $xa_1 + yb_1 \leq xa_2 + yb_2$, $x > 0$ and $a_1 > a_2$, then $b_2 > b_1$.
Proof: $0 \lt x (a_1 - a_2) \leq y (b_2 - b_1)$.
Note: If $ x \geq 0$, then $ b_2 \geq b_1$. The proof is similar.
Back to the problem Case: $ 1 = x \geq -y \geq -z > 0$.
$(x-y) a_{11} + (z-y)a_{13} = -y = (x-y) a_{21} + (z-y)a_{23}. $
Since the coefficients are non-negative and $a_{11} > a_{21}$, hence $a_{23} \geq a_{13}$.
On the other hand,
$ -y a_{22} - za_{23} = xa_{21} < xa_{11} = -ya_{12} -za_{13} $
Since the coefficients are positive and $a_{22} > a_{12}$, hence $a_{13} > a_{23}$.
Thus, we have a contradiction.